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2. If a positive integer n has exactly two positive factors what is the value of n?

Number of factors of a number is \(a+1\) where the number is \(n^a\)And since n has 2 factors\(n\) must be prime and\(>1\)

(1) n/2 is one of the factors of nOnly \(2\) fits these conditions, so \(n=2\)Sufficient

(2) The lowest common multiple of n and n + 10 is an even number.Only \(2\) fits these conditions, so \(n=2\) once again. \(n\) IMO is prime so the only prime that respect statement (2) is 2 because all other prime are odd, and odd+even = odd, so the LCM of an odd and an odd is odd in all cases except n=2

IMO D
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25 Mar 2013, 06:24

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6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negativeNot sufficient Example: \(S = {-1,3,5}\)the product is always <0 but 2 numbers are positiveExample: \(S = {-1,-3,-5}\)the product is always <0 and all numbers are negative

(2) The product of the smallest and largest integers in the list is a prime number.Not sufficientExample: \(S = {1,3,5}\)\(1*5=5\) prime but all positiveExample: S = \({-1,-3,-5}\)\(-1*-5=5\) prime but all negative

(1)+(2) Sufficient IMO CUsing statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative
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------0------------1----------2----------------------|~~~~~~x>y~~~~~~~|~~~x+y<4 for the first one------|~~x=2~~~|~~~~~~x+2y=4~~~~for the second one

And combining all the cases together we obtain1)\(0<x<=2\) with \(0<y<=1\)\(x=2\) and \(x>y\) so \(x=2\) and \(y=1\)2)\(0<x<=2\) with \(y>1\)\(x>y\) and \(x+2y=4\), given that x and y are positive \(x=2, y=1\)3)\(x>2\) with \(0<y<=1\)\(x+y<4\) and \(x=2\) so \(x=2,y=1\)4)x>2 with y>1\(x+y<4\) and \(x+2y=4\), \(x=2,y=1\)In each case \(x=2\) so x is prime
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25 Mar 2013, 07:23

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5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number.the median of three elements is the one in the middle, so b! is oddthere are only 2 cases in which n! is odd and are if n=1 or if n=0so b is 0 , 1 Not Sufficient

(2) c! is a prime numberc can once again be 0,1 or in this case 2.Not sufficient-This is a weak passage, I don't know if I'm right-n! is possible only for positive number so given that a < b < cc must be 2, b must be 1, and (because of my weak hypothesis a>=0) a must be 0

IMO C
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8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime numberNot sufficient

(2) The product of any two terms of the set is a terminating decimalBecause \(\frac{1}{prime}\) is not a terminating decimal, with the only exception of 1/4, 1/1, 1/5 and 1/2any set made by these three CANNOT have a median < 1/5, it can be = 1/5 but NEVER <Some examples:A={1,1,1,1,1/5,1/5,1/5,1/5,1/5,1/5} the median is 1/5 = 1/5A={1,1,1,1,1/2,1/2,1/2,1/2,1/2,1/2} the median is 1/2 > 1/5SUFFICIENT

IMO B
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25 Mar 2013, 08:35

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11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y(2) x + y - 3 = |1-y|

We know that x >0 and y>0 and they are integers.

From F.S 1, we have 2-y>=0 or y<=2. Thus y can only be 2 or 1. Now if y=2, we would have 0>some thing positive or 0>0(when x also equal to 2). Either case is not possible. Thus, y can only be 1. For y=1, we can only have x = 2. Which is prime. Sufficient.

From F.S 2, we have either y>1 or y<1. Now as y is a positive integer, y can't be less than 1.For y=1, we anyways have x=2(prime).In the first case, we have y>1--> x+y-3 = y-1 or x=2(prime). Thus, Sufficient.

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25 Mar 2013, 09:06

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7.Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6(2) When x is divided by 14 the remainder is 2

From F.S 1, we have x = 12q+6 --> 6(2q+1). For x to be a square of an integer, we should have 2q+1 of the form 6^pk^2, where both q,p and k are integers and p is odd. Now we know that 2q+1 is an odd number and 6^pk^2 is even. Thus they can never be equal and hence x can never be the square of an integer. Sufficient.

From F.S 2, we have x = 14q+2 --> 2(7q+1).Just as above, we should have 7q+1 = 2^pk^2. Now for q=1, k=2 and p=1, we have 8=8, thus x is the square of an integer. But for q=0, x is not. Insufficient.

Basically for the second fact statement, we can plug in easily. No need for the elaborate theory.

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25 Mar 2013, 10:02

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5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

(1) The median of {a!, b!, c!} is an odd number(2) c! is a prime number

From F.S 1, we have b! = odd, thus b can be 0 or 1.But, as factorial notation is only for positive integers, thus, if b=0, then a would become negative and thus b is only equal to 1.Now, a can only be 0 as we are given that a! exists. But nothing has been mentioned about c. All we know is that c>1 and an integer. Insufficient.

From F.S 2, we have c! is a prime number. Again, c has to be positive and c can only be 2.However, a and b can take any values, even negative. Insufficient.